7. Common Types

Following are some common type of percentages questions asked in the entrance tests.

7.1 Changing the Base

When one variable ($\bm{x}$) is written as a percentage of another ($\bm{y}$), then changing the base is writing $\bm{y}$ as a percentage of $\bm{x}$. This concept is similar and an extension of section 5. Product Constancy.

Where $\bm{x}$ and $\bm{y}$ are two variables,

- if $\bm{x}$ is $\dfrac{a}{b}$ more than $\bm{y}$, then $\bm{y}$ is $\left( \dfrac{a}{b + a} \right)$ less than $\bm{x}$.

- if $\bm{x}$ is $\dfrac{a}{b}$ less than $\bm{y}$, then $\bm{y}$ is $\left( \dfrac{a}{b - a} \right)$ more than $\bm{x}$.

Example 22

If Aparna has $15\%$ more mangoes than Bhavna, then Bhavna has ___$\%$ less mangoes than Aparna. Fill in the blank.

(1) $11\%$ (2) $13\%$ (3) $17\%$ (4) $19\%$

Solution

Aparna has $\dfrac{15}{100} = \dfrac{3}{20}$ more mangoes than Bhavna.

∴ Bhavna has $\dfrac{3}{20 + 3} = \dfrac{3}{23}$ less mangoes than Aparna

As a percentage, $\dfrac{3}{23} = \dfrac{3}{23} \times$ 100% ~ 13%

Answer: (2) 13%

7.2 Working Backwards

If a certain value, $x$, when increased by $r\%$ results in $y$, then the equation is

$x \times \left( 1 + \dfrac{r}{100} \right) = y$

If we are asked to find $x$, then we write it as
⇒ $ x = y \times \left( \dfrac{100}{100 + r} \right)$

Note that the above need not be memorised. This is simply shown for your understanding. As shown in the example below, key to answering these questions is forming the first equation.

Example 23

A store sells shirts at Rs. $810$, with a profit margin of $8\%$. What is the cost price of a shirt?

Solution

Where CP, SP and $p$ are the Cost Price, Selling Price and Profit $\%$, the formula for calculating SP is as follows.

$\text{CP} \times \left( 1 + \dfrac{p}{100} \right) = \text{SP}$

⇒ $ \text{CP} \times \left( 1 + \dfrac{8}{100} \right) = 810$

⇒ $ \text{CP} = 810 \times \dfrac{100}{108} = 750$

Answer: Rs. $750$

7.3 Linking different bases

In this lesson, we have dealt with elementary examples so far. At CAT level, the questions will have multiple percentage computations like in the example below. Each of these percentages have different bases.

Example 24

In May 2019, Albert spends 80% of his salary. The rest are his savings for the month. In the subsequent month, he receives a 20% increase in his monthly salary but his expenses for the month also increase by 30%. What was the percentage change in his savings in June 2019?

(1) $-10\%$ (2) $-20\%$ (3) $10\%$ (4) $20\%$

Solution

As the question is only in percentage terms, we can assume values. Secondly, for the 2 months, we have expenses and savings. Therefore, we can use a table to fill details.

Let the salary in May 2019 be Rs. 100.
∴ Expenses and Savings in May 2019 are Rs. 80 and Rs. 20 respectively.

	May	June
Expense	80
Savings	20
Salary	100

Salary in June 2019 = $1.2 \times 100 = 120$
Expenses in June 2019 = $1.3 \times 80 = 104$
Savings in June 2019 = 120 - 104 = 16

	May	June
Expense	80	104
Savings	20	16
Salary	100	120

% change in savings in June 2019 = $\dfrac{16 -20}{20} \times$ 100% = - 20%

Answer: (2) –20%

7.4 Dependencies

An example of this concept is “Wealth of a person is being divided amongst 5 children such that A gets 20%, B gets 30% of the remainder, C gets 40% of the remainder, etc.”

As the percentage is always on the remainder, what a subsequent child gets is dependent on what the previous child/children received.

Example 25

Rs. $89,538$ was to be paid up as capital by the $4$ partners - P, Q, R and S. P paid $25\%$ of this amount. Q paid $20\%$ of the remainder. R's capital contribution was $4$ times that of S. If the requirement was exactly covered by these capital contributions, then what was the capital contribution of S?

(1) Rs. $8,754$ (2) Rs. $10,745$ (3) Rs. $12,342$ (4) Rs. $15,679$

Solution

The total capital contribution of Rs. $89,538$ is a difficult figure to work with. Therefore, let's assume that the total capital is Rs. $100$.

P's capital = $\dfrac{25}{100} \times 100 = 25$

Q's capital = $\dfrac{20}{100} \times (100 - 25) = 15$

Capital of R and S = $100 - 25 - 15$ = Rs. $60$

Let the capital contribution of S be $x$. Then, that of R = $4x$.

$4x + x = 60$
⇒ $ x = 12$

Where the total capital is Rs. $100$, S's contribution is Rs. $12$, which means S's contribution is $12\%$ of the total.

Capital brought in by S = $\dfrac{12}{100} \times 89538$

As the answer options are quite apart, we can round off 89538 to 90000.
$\dfrac{12}{100} \times 89538 \sim \dfrac{12}{100} \times 90000 = 10800$

The closest answer here is Rs. $10,745$.

Answer: (2) Rs. $10,745$

7.5 Relationship between percentages

The following is an example of percentages being used where variables are related to each other.

Example 26

A class has $5$ students and all of them wrote a Science test. The test had $2$ parts - Physics and Chemistry. Rahul scored $30\%$ of the total marks scored by the entire class. $20\%$ of Rahul's marks came from Chemistry. If Rahul's marks in Physics was $40\%$ of the total marks scored by the entire class in Physics, then the ratio of marks scored in Chemistry by Rahul to that of the rest of the class is

(1) $3 : 17$ (2) $4 : 21$ (3) $7 : 23$ (4) $19 : 81$

Solution

With no absolute values given, we can assume 100 as total marks scored by all students in the Science test.

Secondly, Rahul's marks are provided as a percentage of the entire class. As Physics and Chemistry marks are there for each, we can use a table to fill details.

Rahul's total marks = $\dfrac{30}{100} \times 100 = 30$

Rahul's marks in Chemistry = $\dfrac{20}{100} \times 30 = 6$

	Physics	Chemistry	Total
Rahul	24	6	30
Rest of the Class			70
Total			100

Rahul's marks in Physics is 40% of that of the entire class.

∴ Total Physics marks = $24 \times \dfrac{100}{40} = 60$

∴ Total Chemistry marks = 100 – 60 = 40

	Physics	Chemistry	Total
Rahul	24	6	30
Rest of the Class	36	34	70
Total	60	40	100

Ratio of Rahul's marks in Chemistry to that of the entire class = 6 : 34 = 3 : 17

Answer: (1) $3 : 17$